Question

The area bounded by the curve $y=f\left(x\right)$, x-axis and the ordinates $x=1$ and $x=b$ is $(b-1)\sin (3b+4)$, then $f\left(x\right)$ equals-

• A. $(x-1)\cos (3x+4)$
• B. $\sin (3x+4)$
• C. $\sin (3x+4)+3(x-1)\cos (3x+4)$
• D. $Noneofthese$

Answer 1

The correct option is C $\sin (3x+4)+3(x-1)\cos (3x+4)$

$\mathrm{Re}quiredarea$
$\underset{1}{\overset{b}{\int }}f\left(x\right)dx=(b-1)\sin (3b+4)$
$differentiatingbothsidesw.r.t.b$
$f\left(b\right)=(b-1)3\cos (3b+4)+\sin (3b+4)$
$=3(b-1)\cos (3b+4)+\sin (3b+4)$
$atb=x$
$f\left(x\right)=3(3x-1)\cos (3x+4)+\sin (3x+4)$
Option C is correct answer.